Problem: Simplify the following expression: $y = \dfrac{-5x^2- 12x- 4}{-5x - 2}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-5)}{(-4)} &=& 20 \\ {a} + {b} &=& &=& {-12} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $20$ and add them together. The factors that add up to ${-12}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-2}$ and ${b}$ is ${-10}$ $ \begin{eqnarray} {ab} &=& ({-2})({-10}) &=& 20 \\ {a} + {b} &=& {-2} + {-10} &=& -12 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-5}x^2 {-2}x) + ({-10}x {-4}) $ Factor out the common factors: $ x(-5x - 2) + 2(-5x - 2)$ Now factor out $(-5x - 2)$ $ (-5x - 2)(x + 2)$ The original expression can therefore be written: $ \dfrac{(-5x - 2)(x + 2)}{-5x - 2}$ We are dividing by $-5x - 2$ , so $-5x - 2 \neq 0$ Therefore, $x \neq -\frac{2}{5}$ This leaves us with $x + 2; x \neq -\frac{2}{5}$.